# Confused about vectors into rectangular components

So, i have been learning about forces, vectors and such. I know many times for convience we break vector forces into rectangular components...fx=Fcostheta, fy=Fcostheta. It is easier to do regular alegbra verus vector alegrba

My question is both fx and fy are scalars, but then they can be negative depending on the angle. How so, arent scalars magnitude without direction, and isnt magnitude a positive value only? Or is magnitude only positive for vectors.....scalars can be either?

I see we work with force as a scalar values to represent vectors, can someone please explain this? Just confused is all. Thanks

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rock.freak667
Homework Helper
So, i have been learning about forces, vectors and such. I know many times for convience we break vector forces into rectangular components...fx=Fcostheta, fy=Fcostheta. It is easier to do regular alegbra verus vector alegrba

My question is both fx and fy are scalars, but then they can be negative depending on the angle. How so, arent scalars magnitude without direction, and isnt magnitude a positive value only? Or is magnitude only positive for vectors.....scalars can be either?

I see we work with force as a scalar values to represent vectors, can someone please explain this? Just confused is all. Thanks
When you resolve vectors, you end up with two vectors. One ends up being horizontal and the other vertical. This is why they can be positive or negative to indicate direction. So if you have a vector of magnitude F acting at an angle of θ to a horizontal, then

Fx = Fcosθ and Fy= Fsinθ with Fx being in a horizontal direction and Fy being in a vertical direction.

Okay i got that u break the vector into to other vectors, but arent fx and fy still scalars? Arent scalars defined as value without direction, can they be negative and positive them?

Are the values of fx and fy scalar magnitude of force where the negative just represents direction?

haruspex
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Arent scalars defined as value without direction, can they be negative and positive them?
Scalars still have sign, just as +1 is different from -1. What's confusing you is that there's an ambiguity in representation of a vector as a scalar multiplied by a unit vector: if you reverse the vector and reverse the sign on the scalar you're back to the same vector.

My question is both fx and fy are scalars, but then they can be negative depending on the angle. How so, aren't scalars magnitude without direction, and isn't magnitude a positive value only? Or is magnitude only positive for vectors.....scalars can be either?
-1 is a scalar. It is also a vector, because all scalars are vectors, but not all vectors are scalars. Math is fun.

Scalars are numbers that "scale" vectors. Think of them as lengthening or shortening them. If the scalar is negative, it reverses the direction of the vector (unless it's a zero vector). Think of it this way:

If I walk y meters north and x meters east, then what happens if I move y meters south and x meters west? It's in the opposite direction. You can think of the second one as moving -y meters north and -x meters east.

Okay, can someone check my understanding. So what we are doing is breaking down a vector into rectangular components which is easier. Then we can use scalars with negative and postive values to represent the vectors? Is that correct? The scalars fx and fy are scalar numbers, that can be neg or positive and represent the direction of the vectors

-1 is a scalar. It is also a vector, because all scalars are vectors, but not all vectors are scalars. Math is fun.

Scalars are numbers that "scale" vectors. Think of them as lengthening or shortening them. If the scalar is negative, it reverses the direction of the vector (unless it's a zero vector). Think of it this way:

If I walk y meters north and x meters east, then what happens if I move y meters south and x meters west? It's in the opposite direction. You can think of the second one as moving -y meters north and -x meters east.
Ok, this might confuse me more, can u please just help me understand how rectangular components work with negative and positive values, representing forces. That is what i am concerned about. Thanks

SteamKing
Staff Emeritus
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You are not understanding that the components of a vector are vectors themselves. These component vectors have directions which are parallel to the x, y, or z coordinate axes. In order to simplify things, we use vectors which have a magnitude of 1, which are also called unit vectors. The unit vector in the x-direction is usually denoted by a lower case i with a ^ over it (also called 'i-cap'; the unit vector in the y-direction is called 'j-cap', and 'k-cap' for the z-direction. These unit vectors also have a direction angle of 0, which means they are pointing in the positive direction of their respective coordinate axes.

Unit vectors can be multiplied by scalar values. If the scalar value is positive, then the magnitude of the unit vector increases, but the angle between the vector and the coordinate axis is still 0. If the scalar value is negative, the magnitude of the vector increases, but the angle is increased by pi radians to indicate that the new vector is pointing in the opposite direction.

haruspex
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all scalars are vectors
Not sure I go along with that. Yes, you can set up a bijection between scalars and one dimensional vectors that preserves a lot of structure, but that's not the same as saying scalars are one dimensional vectors. In the definition of a vector space, there is no product function VxV→V, but there is in the underlying scalar field.

Umm, i get unit vectors. I get that Fx=fxi and Fy=fyi are vectors and the scalar components are fx and fy. I am jsut curious what it means when the scalar is negative.

For example, say i have a vector FX=-50Ni, what does the scalar component -50N mean? Does it mean it is a 50 newton force, opposite direction?

Also, scalars are always define as just magnitude. Yet magnitude is always an absolute value, yet scalars can be both? Maybe me defintio n is too general.

Its good to think of what your positive directions are first. Usually, [up] and
are positive direction notations. There fore, using this convention, a vector that is - will be up left or down (depends on the component you are analysing). So a scalar quantity that is negative represents a quantity that is -, really that's it.​

haruspex
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scalars are always define as just magnitude. Yet magnitude is always an absolute value
The mathematical definition of a vector space is as a "group over a field". This means e.g. that you can add and subtract the vectors and multiply them by elements of the field (being the scalars); the result in each case being other vectors. A field supports addition, subtraction, multiplication and division (except by 0) within itself. In particular:
- the field has an additive identity, that is, an element (usually denoted 0) such that for any x in the field x+0=x.
- for any x in the field, there is also an element (written -x) such that x + (-x) = 0
So, by definition, scalars have a concept of sign.
Typically, the field used for a vector space is either the real numbers or the complex numbers. In each case, there is also a concept of magnitude, but that does not capture all the details of the scalar.

Okay, so my explanation was right. When we break down vectors, we can break them down into thier scalar components....fx=Fcostheta and fy=Fsintheta. These values pertain to the forces and if they are positive or negative, that means the force has a down or up direction depending on how the coordinate plane is defined correct?

haruspex
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Okay, so my explanation was right. When we break down vectors, we can break them down into thier scalar components....fx=Fcostheta and fy=Fsintheta. These values pertain to the forces and if they are positive or negative, that means the force has a down or up direction depending on how the coordinate plane is defined correct?
Sounds fair.

gneill
Mentor
Okay, so my explanation was right. When we break down vectors, we can break them down into thier scalar components....fx=Fcostheta and fy=Fsintheta. These values pertain to the forces and if they are positive or negative, that means the force has a down or up direction depending on how the coordinate plane is defined correct?
You shouldn't get the impression that vectors have scalar components; Vectors have vector components. The scalars involved merely multiply (scale) unit vectors so as to yield the vector components that lie along the given coordinate system axes. The unit vectors are associated with the coordinate system axes, and are parallel to and directed towards the positive direction of the given axis. A negative scalar multiplier has the effect of reversing the sense (direction) of the result so that components can be directed in the negative direction along the given axis, too.

Granted, the procedure of breaking down a vector into its components involves determining these scalar multipliers. But the resulting vector components are these scalar values multiplied by their associated unit vectors. The scalars alone are essentially meaningless without the unit vectors of the coordinate system to orient them in space and thus making them vector components.

I guess i have been concerned about scalar rectangular components because that is what my book does. It takes some force/vector, draws a freebody diagram to picture the all the component vectors. Then, using the equations Fx=Fcostheta and Fy=Fsintheta, they break them down, ignoring the vectorial component and unit vectors. At this point, they are simply using the scalar values to represent the vectors as they do the mathh. Once complete, they then bring it back into a polar notation.

This is why i asked. This is why i assumed that the negative values pertained to a downward sense of direction

Not sure I go along with that. Yes, you can set up a bijection between scalars and one dimensional vectors that preserves a lot of structure, but that's not the same as saying scalars are one dimensional vectors. In the definition of a vector space, there is no product function VxV→V, but there is in the underlying scalar field.
Isn't the real number line a well-defined vector space?

haruspex
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Isn't the real number line a well-defined vector space?
The standard definition does not define it in a way that makes it a vector space. You can construct a 1 dimensional vector space over it and a homomorphism between the two, but that does not mean it is that vector space. For example, the homomorphism is not unique. You could construct the mapping f:ℝ→V as f(x) = (2x), and the dot product °:VxV→ℝ as (x)°(y) =xy/4.

Edit: Scrap the bit about dot product. A vector space doesn't have to have one.

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that does not mean it is that vector space
Well, this is really a question of how one interprets 'is'. Any field is a vector space over itself, because it satisfies all the vector space axioms trivially.

So real numbers are both scalars and vectors.

gneill
Mentor
So real numbers are both scalars and vectors.
So what's the direction of 2.7, and how does it differ from the direction of -113.987? Vectors have both magnitude and direction...

So what's the direction of 2.7, and how does it differ from the direction of -113.987? Vectors have both magnitude and direction...
The directions are 1 and -1, respectively, assuming the natural scalar product in the set of real numbers.

haruspex
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Well, this is really a question of how one interprets 'is'. Any field is a vector space over itself, because it satisfies all the vector space axioms trivially.

So real numbers are both scalars and vectors.
From any field F = (S, +, ×) you can construct a vector space over itself in a natural way. But as I showed, there are also less natural ways to do it.

From any field F = (S, +, ×) you can construct a vector space over itself in a natural way.
'Construct' has a connotation that given X, you also have Y, and X is not necessarily Y. However, in the natural case, X is Y exactly.

But as I showed, there are also less natural ways to do it.
You had to introduce operations that were not present in the original field, which makes the result a different object. But one can always turn something into something else in this way, and I cannot see what this sort of argument can possibly prove.